Trigonometry/Circles and Triangles/The Excentral Triangle

Let ABC be a triangle, its incentre be I and its three excentres be Ia, Ib and Ic. Then IaIbIc is the excentral triangle of ABC.

A lies on the line IbIc and is the foot of the perpendicular from Ia to that line, and similarly for B and C. Thus ABC is the pedal triangle (see later) of its excentral triangle. Further, these perpendiculars intersect at I, so I is the orthocentre of the excentral triangle.

I and Ia lie on the bisector of angle BAC and similarly for the other excentres.

The quadrilaterals IBCIa, IACIb and IABIc are cyclic.

Ia is the orthocentre of the triangle IbIcI and similarly for the other excentres.

The distance IIa is 4Rsin($A/2$) and similarly for the other excentres.

Properties of the Excentral Triangle
The angles of this triangle are $$90^0-\frac{A}{2}, 90^0-\frac{B}{2} \text{ and } 90^0-\frac{C}{2}$$.

Its sides are $$4R\cos\left(\frac{A}{2}\right), 4R\cos\left(\frac{B}{2}\right), 4R\cos\left(\frac{C}{2}\right)$$.

Its area is $$8R^2\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right) = 2Rs = \left(\frac{1}{2}\right) \Delta \text{cosec}\left(\frac{A}{2}\right) \text{cosec}\left(\frac{B}{2}\right) \text{cosec}\left(\frac{C}{2}\right)$$

where &Delta; is the area of the original triangle.

Its circumradius is 2R.